21sorry for the post above. i will repair that
$\textbf{ If $\mathbf{\alpha,\beta,\gamma}$ be the Roots of the equation $\mathbf{x^3+2x^2-3x-1=0}$.\\\\ Then find the value of $\mathbf{\frac{1}{\alpha^5}+\frac{1}{\beta^5}+\frac{1}{\gamma^5}=}$}
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21The best way will be to find an equation with roots as 1/α5,1/β5,1/γ5
which can be done by replacing x by _________
then we can collect the co efficient of x____? as reqiured answer.
1dude i did the same ,u end up getting poynomial of degree 5 which will be having 5 roots
,these 3 roots +2 unwanted roots
21The polynomial with roots as a= 1/α , b = 1/β, c=1/γ is given by y3+ 3y2-2y -1 = 0
we have a + b + c = - 3 and ab + bc + ac = 2 which yields a2+b2+c2 = 5
We will denote sum of n-th square of roots as Sn. (a+b+c = S1)
a,b,c are the roots of y3+ 3y2-2y -1 = 0
therefore a3+ 3a2-2a -1 = 0
b3+ 3b2-2b -1 = 0
c3+ 3c2-2c -1 = 0
on addition, S3 + 3S2 -2S1 = 3, which gives S3 = -18
Consider the equation y( y3+ 3y2-2y -1) = 0 we have just introduced zero as a root. So Sn's will not change.
proceeding on above way we get S4+ 3S3 - 2S2 - S1= 0, so S4= 69
From the equation y2( y3+ 3y2-2y -1) = 0 we get S5+ 3S4-2S3 - S2 = 0 which gives S5 = -238
So, a5+ b5+ c5= 1/α5+ 1/β5+ 1/γ5 = -238
